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JAMB Physics 2018 Paper

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Question : 74

Total: 87

A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms

^{-1}

. Determine the magnitude of the resulting impulse

20.0kgms $^{-1}$
10.0kgms $^{-1}$
5.0kgms $^{-1}$
2.5kgms $^{-1}$

Solution:

Concept:

Impulse is equal to the change in momentum of the ball.

Explanation:

For a smooth vertical wall hit normally, the ball rebounds with the same speed in the opposite direction.

Let initial velocity be

u = +2\,\mathrm{m/s}

.

Then final velocity is

v = -2\,\mathrm{m/s}

.

Impulse

= m(v - u)

.

= 5(-2 - 2)

= 5(-4) = -20\,\mathrm{kg\,m/s}

.

The magnitude of the impulse is therefore

20.0\,\mathrm{kg\,m/s}

.

Answer:

A.

20.0\,\mathrm{kg\,ms^{-1}}

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